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# Python | Get Top N elements from Records

• Last Updated : 27 Jan, 2023

Sometimes, while working with data, we can have a problem in which we have records and we require to find the highest N scores from it. This kind of application is popular in web development domain. Let’s discuss certain ways in which this problem can be solved.
Method #1 : Using sorted() + lambda The combination of above functionality can be used to perform this particular task. In this, we just employ sorted function with reverse flag true, and print the top N elements using list slicing.

## Python3

 `# Python3 code to demonstrate working of``# Get Top N elements from Records``# Using sorted() + lambda` `# Initializing list``test_list ``=` `[(``'Manjeet'``, ``10``), (``'Akshat'``, ``4``), (``'Akash'``, ``2``), (``'Nikhil'``, ``8``)]` `# Initializing N``N ``=` `2` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# Get Top N elements from Records``# Using sorted() + lambda``res ``=` `sorted``(test_list, key ``=` `lambda` `x: x[``1``], reverse ``=` `True``)[:N]` `# printing result``print``(``"The top N records are : "` `+` `str``(res))`

Output

```The original list is : [('Manjeet', 10), ('Akshat', 4), ('Akash', 2), ('Nikhil', 8)]
The top N records are : [('Manjeet', 10), ('Nikhil', 8)]
```

Method #2 : Using sorted() + itemgetter() The combination of above functions can also be used to perform this particular task. In this, the task performed by lambda function is performed by itemgetter() is used to get the index in tuple which has to be included in calculations.

## Python3

 `# Python3 code to demonstrate working of``# Get Top N elements from Records``# Using sorted() + itemgetter()``from` `operator ``import` `itemgetter` `# Initializing list``test_list ``=` `[(``'Manjeet'``, ``10``), (``'Akshat'``, ``4``), (``'Akash'``, ``2``), (``'Nikhil'``, ``8``)]` `# Initializing N``N ``=` `2` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# Get Top N elements from Records``# Using sorted() + itemgetter()``res ``=` `sorted``(test_list, key ``=` `itemgetter(``1``), reverse ``=` `True``)[:N]` `# printing result``print``(``"The top N records are : "` `+` `str``(res))`

Output

```The original list is : [('Manjeet', 10), ('Akshat', 4), ('Akash', 2), ('Nikhil', 8)]
The top N records are : [('Manjeet', 10), ('Nikhil', 8)]
```

Method #3 : Using heapq.nlargest()
This approach uses the heapq library’s nlargest() function, which returns the top N elements from an iterable. This approach is more efficient than the previous two, as it uses a heap data structure to keep track of the largest elements.

## Python3

 `import` `heapq` `# Initializing list``test_list ``=` `[(``'Manjeet'``, ``10``), (``'Akshat'``, ``4``), (``'Akash'``, ``2``), (``'Nikhil'``, ``8``)]` `# Initializing N``N ``=` `2` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# Get Top N elements from Records``# Using heapq.nlargest()``res ``=` `heapq.nlargest(N, test_list, key ``=` `lambda` `x: x[``1``])` `# printing result``print``(``"The top N records are : "` `+` `str``(res))``#This code is contributed by Edula Vinay Kumar Reddy`

Output

```The original list is : [('Manjeet', 10), ('Akshat', 4), ('Akash', 2), ('Nikhil', 8)]
The top N records are : [('Manjeet', 10), ('Nikhil', 8)]
```

The time complexity of this approach is O(n log n) as it uses a heap data structure which is logarithmic in nature and Auxiliary space is O(n) as it stores the top N elements in memory.

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